User blog:Mh314159/A hopefully powerful new system
Please comment on my latest attempt to climb toward TREE(3) as a personal intellectual challenge. Conceptually, here's how this works. The base function is successorship. Then there is a line of bracket-enclosed numbers or comma-separated strings of numbers. The last two brackets expressions and the "rank" and "argument". All others to their left are "orders". There are rules that recurse longer and more complex expressions down to simpler expressions. A number to the left of a backslash indicates the cycle of the expressions to its right and there is a powerful cycle factorial rule. In order from most complex to least complex, expressions recurse in the following order of rule sets: c > o > ss > sn > ns > nn > base. I think this generates numbers much bigger than anything else I have tried to do. I am not necessarily done with this. Here are the rule sets with some examples. For two sets of brackets, the number or string in the left brackets is the "rank" of the function and the number or string in the right brackets is the "argument" Rule set n (single bracket containing a number, not part of a larger expression on the same level (including superscript or subscript) a = a (this is called a "terminated number") Rule set nn (two bracket sets contain numbers) n1 = n+1 1a = a+1 na = n-1na-1a functional recursion notation Some examples: 21 = 3 = 3 22 = 132 = 5 23 = 1223 = 153 = 8 24 = 1234 = 184 = 12 25 = 1245 = 1125 = 17 26 = 1256 = 1176 = 23 2n = 3 + 2 + 3 + 4 + ... n = 3 + (n-1)(n+2)/2 2n grows approximately as fast as (n^2)/2 31 = 4 32 = 242 = 12,092 33 = 2323 = 212,0923 = approx 3 repeatedly squared 12,092 times, so 3^(2^12,092) which is greater than 3^2^3^8 which is greater than 3^^4 34 = 2334 > 23^^44 > 4 repeatedly squared 3^^4 times, so > 4^(2^(3^^4)) or approx 3^^6 35 = 2345 > 3^^8 3n > 3^^(2n-2) 41 = 5 42 = 352 = 333332 > 33333^8 = 333(3^^3^8) = 33(3^^3^^3^8) > 333^^^3 > 3(3^^(3^^3^^3) = 3(3^^^4) > 3^^^5 43 > tetrating 3 3^^^5 times, > 3^^^6 4n = approx n^^^(n+3) nn > n^...^n with n-1 up-arrows Rule set ns bracket contains number, right bracket contains string) drop all zeroes and collapse remaining entries y is a string of two more entries and final entry q (i.e., y = a,b,...,q), y' is y with the penultimate entry decreased by 1. [ny = [ny'] [ay',by',...q-1] 12,2 = [11,2][21,2,1] 11,2 = [12][12,1] = 44,1 44,1 = [43,1][43,1] 43,1 = [42,1][42,1] 42,1 = [41,1][41,1] 41,1 = [41][41] = 55 so if f(n) = nn, then nn,1 is f nn+1 44,1 corresponds approximately to G4, the fourth iteration in Graham's sequence 21,2 = [22][12,1] = 53,1 > G3 31,2 = [32][32,1] = 1209212092,1 > G12092, the 12092nd iteration of Grahams' sequence on which G64 = the Graham-Gardner number, and is therefore the smallest expression in this system that is greater than that famous large number. 12,2 = [11,2][21,2,1] = [44,1][21,2,1] = G4G3,1 is therefore greater than G3 iterations of Graham's sequence 22,2 = [21,2][21,2,1] 32,2 = [31,2][31,2,1] > 31,2 iterations of Graham's sequence and therefore much larger than Conway's 3-->3-->3-->3, which is G27 iterations. 12,1,2 = [12,2][22,2,12,2,1] because the second term in the string is a large number, this is incomparably larger than all of the previous numbers. 22,1,2 = [22,2][22,2,12,2,1] 12,2,2 = [12,1,2][22,1,2,22,1,2,1] Rule set sn (left bracket contains string, right bracket contains number) drop all zeroes and collapse remaining entries y is a string of two more entries a,b,...q y' is the y with the penultimate entry decreased by 1 yn = [ny']n,n,...,n instances of n = [ay',by',...,q-1]n Rule set ss (left bracket contains string, right bracket contains string) Define x = string of two or more entries Define y = string of two or more entries Define y' = string y with its penultimate entry decreased by 1 Define z = string y with its last entry decreased by 1 and all other entries replaced by xy' drop zeroes and collapse expressions xy = x,x,...xz with xz instances of x Example: 2,3,45,6,7 = 2,3,4,2,3,4,...[2,3,45,5,7,2,3,45,5,7,6] with 2,3,4[2,3,45,5,7,2,3,45,5,7,6] instances of 2,3,4 From the previous expression, the number of groups of 2,3,4 in the elliptical string would recurse as follows: 2,3,4[2,3,45,5,7,2,3,45,5,7,6] = 2,3,4,2,3,4,...[2,3,4y',2,3,4y',5] with 2,3,4[2,3,4y',2,3,4y',5] instances of 2,3,4 and where y' = 2,3,45,5,7,(2,3,45,5,7-1),6 and the next step would be to begin to find the value of 2,3,45,5,7 We would begin to recurse the simplest possible string/string expression as follows: 1,11,1 = 1,1,...[1,11] with 1,1[1,11] 1's 1,11 = [11]1,1,... with [11]1 = 21 = 3 1's 1,11 = [11]1,1,1 = 21,1,1 21,1,1 = [21,1][11,1,11,1] 11,1 = [11][11] = 22 = 5 21,1,1 = [21,1]5,5 21,1 = [21][11] = 12092 21,1,1 = 120925,5 1,11,1 = 1,1,...[120925,5] with 1,1[120925,5] 1's Rule set o for n sets of expressions Define x, y, z to be numbers or strings of numbers Define z' = string z with its penultimate entry decreased by 1 Define z'' = string z with every term except the last replaced by xyz' and the last term decreased by one. drop trailing zeroes if z is a single number, xy...z (any number n of consecutive bracketed expressions) = x,x,...xy,y,...y...z-1 with xy...z-1 instances of x, y, etc. in each of the first n-1 sets of expressions if z is a string, xy...z = x,x,...xy,y,...y...z'' with xyz' instances of x, y, etc. in each of the first n-1 sets of expressions 12,34 = 1,1,...2,3,2,3,...3 with 12,33 instances in each of the first two sets of expressions 12,32,2 = 1,1,...2,3,2,3,...[12,31,2,1] with 12,31,2 instances in the first two sets of expressions Eventually, the last expression develops zeroes and drops them and recurses to nothing while the remaining strings become long; when there are only two sets of brackets, the ss rules apply. Rule set c The cycle rule triggers when an expression is of the simple form p\a where a is a number. It also triggers when there is an expression of the form p\N and during the recursing of N, an expression of the type n1 or 1a is reached and then the cycle number distributes past the backslash and is used to evaluate that expression. When clarity is needed for embedding a recursed expression in an expression with a higher cycle number, the 0\ indicator should be retained, indicating a terminated number to which the cycle rule will not further apply. Or, the recursed expression can be set equal to a single unbracketed variable which can then be used in the original expression. Lower case letters stand for terminated numbers and upper case letters stand for bracketed expressions. Notice that in expressions of the form p\N where N is an expression containing superscripts that are themselves bracketed expressions, the superscripts are not functional recursion numbers until after they have been fully recursed by the cycle rule. p\1 = p+1 for a single number a, p\a = p-1\a,a,...,a...a,a,...,a with p-1\a sets of brackets and p-1\a terms in each string. p\1a = p\u where u = p-1\v etc., terminating at z = 0\1a Equivalently, p\1a = p\[p-1\[p-2\...[0\1a]]] Example: 3\12 = 3\[2\[1\[0\12]]] p\a1 = p\u where u = p-1\v etc., terminating at z = 0\a1 Examples: 1\32 = 1\2312 and by the cycle rule 1\31 = 1\u where u = 0\31 = 4 and 1\4 = 0\4,4,4,44,4,4,44,4,4,44,4,4,4 and if this is m, then 1\32 = 1\2m2 = 1\22...22 with m instances of 2 and the last instance is evaluated as 1\22= 1\1212and a similar process is used to turn this into a string of 3,3,33,3,33,3,3 1 functions, the rightmost of which is evaluated as 1\12 = 0\3,3,33,3,33,3,3 and if this terminated number is p, then the next function to the left (of a very long sequence) is 1\1p 2\11 = 2\u where u = 1\v where v = 0\11 = 2, so u = 1\2 = 0\2,22,2and therefore 2\1,1 = 2\p where p is the large number 2,22,2 = and 2\p = 1\p,p,p...p,p,p with 1\p sets of brackets and 1\p terms in each bracket. This is then evaluated using rules o through nn and the cycle rule down to a cycle zero number. Alternatively, 2\11 = 2\[1\[0\11]] = 2\[1\2] = 2\P where P is 0\2,22,2 which is a terminated number p and therefore 2\1,1 = 2\p = 1\p,p,p...p,p,p with 1\p sets of brackets and 1\p terms in each bracket. Once 1\p is evaluated then the number of brackets and terms is know and 1\p,p,p...p,p,p is then evaluated using rules o through nn and the cycle rule down to a cycle zero number. 2\15 = 2\u with u = 1\v with v = 0\15 = 6. Therefore u = 1\6 = 0\M where M is an expression of 6 consecutive brackets each containing a string of six 6's. This is a cycle zero and therefore u is a terminated number, and so 2\u = 1\P where P is an expression of 1\u consecutive brackets each containing a string of 1\u u's. The next step is to recurse 1\u to cycle zero to know the number of brackets and string lengths in expression P. Once this is known, then 1\P can be recursed to cycle zero. Equivalently, 2\15 = 2\[1\[0\15]] = 2\[1\[0\6]] = 2\[1\6] = 2\u = 1\P where P is 1\u consecutive brackets each containing 1\u u's. Expressions containing multiple backslashes associate from left to right, so that c\d\e = (c\d)\e 1\2\2 = N\2 where N = 2,22,2 and then the p\a rule would apply, so the next recursion is N\2 = N-1\2,2,...,2...2,2,...,2 with N-1\2 brackets and terms per string. Many more recursions have to be performed in order to finally determine the number of brackets in N\2. Category:Blog posts